SOLUTION: 'A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were: The fifth number plus the thi

Algebra ->  Equations -> SOLUTION: 'A man wanted to get into his work building, but he had forgotten his code. However, he did remember five clues. These are what those clues were: The fifth number plus the thi      Log On


   

Question 151039: 'A man wanted to get into his work building, but he had forgotten his code.
However, he did remember five clues. These are what those clues
were:
The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is one less than twice the second number.

The second number plus the third number equals ten.

The sum of all five numbers is 30.'

What were the five numbers and in what order?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let the code be ABCDE.
1.E%2BC=14
2.D=B%2B1
3.A=2B-1
4.B%2BC=10
5.A%2BB%2BC%2BD%2BE=30
Let's see if we can get each number in terms of B.
From 3,
A=2B-1
From 4,
B%2BC=10
C=10-B
From 2,
D=B%2B1
From 1 and 4,
E%2BC=14
E=14-C
E=14-%2810-B%29
E=14-10%2BB
E=B%2B4
Now substitute all of these into 5 and solve for B.
5.A%2BB%2BC%2BD%2BE=30
%282B-1%29%2BB%2B%2810-B%29%2B%28B%2B1%29%2B%28B%2B4%29=30
B%282%2B1-1%2B1%2B1%29%2B%28-1%2B10%2B1%2B4%29=30
4B%2B14=30
4B=16
B=4
Now go back and find A,C,D,and E.
A=2B-1=2%284%29-1=8-1=7
C=10-B=10-4=6
D=B%2B1=4%2B1=5
E=B%2B4=4%2B4=8
His number was 74658.
.
.
.
The question to ask is how could he remember these find intricate codes but not 5 numbers.